Calculate the Standard Deviation and coefficient of variation for the given frequency table:
Class-interval Frequency
$1 - 5$ $1$
$6 - 10$ $2$
$11 - 15$ $3$
$16 - 20$ $4$

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Solution

C.I $f$ Mid point $x$ $f x$ $D = X - ¯ ¯¯¯ ¯ X$ $D^{2}$ $f D^{2}$
$1 - 5$ $1$ $3$ $3$ $- 10$ $100$ $100$
$6 - 10$ $2$ $8$ $16$ $- 5$ $25$ $50$
$11 - 15$ $3$ $13$ $39$ $0$ $0$ $0$
$16 - 20$ $4$ $18$ $72$ $+ 5$ $25$ $100$

$N = 10$
$\sum f x = 130$
$\sum f D^{2} = 250$
Standard Deviation $= σ = \sqrt{\frac{\sum f D^{2}}{N}} = \sqrt{\frac{250}{10}} = \sqrt{25}$
Therefore, $σ = 5$
Coefficient of variation, C.V $= \frac{σ}{¯ ¯ ¯ x} \times 100 = \frac{5}{13} \times 100 = \frac{500}{13} = 38.46$

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$C . I .$	$f$
$1 - 5$ $6 - 10$ $11 - 15$ $16 - 20$	$1$ $2$ $3$ $4$

N = 10

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Calculation (i) Arithmetic Mean (ii) Standard Deviation and (iii) Co-efficient of variation for the following frequency distribution.

Class Interval

Frequency

30 – 35

35 – 40

40 – 45

45 – 50

50 – 55

Total

Class Interval	1−5	6−10	11−15	16−20	21−25	26−30	31−35
Frequency	2	8	15	35	20	10	14

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